pearson chapter 4 homework answers

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0.10 + 0.05 = 0.15

0.35 + 0.40 + 0.10 = 0.85

1(0.15) + 2(0.35) + 3(0.40) + 4(0.10) = 0.15 + 0.70 + 1.20 + 0.40 = 2.45

Let X = the number of events Javier volunteers for each month.

1 – 0.05 = 0.95

0.2 + 1.2 + 2.4 + 1.6 = 5.4

The values of P ( x ) do not sum to one.

Let X = the number of years a physics major will spend doing post-graduate research.

1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15

1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6 years

X is the number of years a student studies ballet with the teacher.

0.10 + 0.05 + 0.10 = 0.25

The sum of the probabilities sum to one because it is a probability distribution.

− 2 ( 40 52 ) + 30 ( 12 52 ) = − 1.54 + 6.92 = 5.38 − 2 ( 40 52 ) + 30 ( 12 52 ) = − 1.54 + 6.92 = 5.38

X = the number that reply “yes”

0, 1, 2, 3, 4, 5, 6, 7, 8

X = the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.

X = the number of business majors in the sample.

2, 3, 4, 5, 6, 7, 8, 9

0, 1, 2, 3, 4, …

X = the number of U.S. teens who die from motor vehicle injuries per day.

0, 1, 2, 3, 4, ...

The variable of interest is X , or the gain or loss, in dollars.

The face cards jack, queen, and king. There are (3)(4) = 12 face cards and 52 – 12 = 40 cards that are not face cards.

We first need to construct the probability distribution for X . We use the card and coin events to determine the probability for each outcome, but we use the monetary value of X to determine the expected value.

  • Expected value = ( 6 ) ( 6 52 ) + ( 2 ) ( 6 52 ) + ( − 2 ) ( 40 52 ) = – 32 52 Expected value = ( 6 ) ( 6 52 ) + ( 2 ) ( 6 52 ) + ( − 2 ) ( 40 52 ) = – 32 52
  • Expected value = –$0.62, rounded to the nearest cent
  • If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.
  • You should not play this game to win money because the expected value indicates an expected average loss.
  • $200,000; $600,000; $400,000
  • third investment because it has the lowest probability of loss
  • first investment because it has the highest probability of loss
  • second investment

Let X = the amount of money to be won on a ticket. The following table shows the PDF for X .

Calculate the expected value of X .

0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35

A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.

X = the number of patients calling in claiming to have the flu, who actually have the flu.

X = 0, 1, 2, ...25

  • X = the number of DVDs a Video to Go customer rents
  • X = number of questions answered correctly
  • X ~ B ( 32,  1 3 ) ( 32,  1 3 )
  • We are interested in MORE THAN 75% of 32 questions correct. 75% of 32 is 24. We want to find P ( x > 24). The event "more than 24" is the complement of "less than or equal to 24."
  • Using your calculator's distribution menu: 1 – binomcdf ( 32,  1 3 ,  24 ) ( 32,  1 3 ,  24 )
  • P ( x > 24) = 0
  • The probability of getting more than 75% of the 32 questions correct when randomly guessing is very small and practically zero.
  • X = the number of college and universities that offer online offerings.
  • 0, 1, 2, …, 13
  • X ~ B (13, 0.96)
  • P ( x = 12) = 0.3186 P ( x = 13) = 0.5882 More likely to get 13.
  • X = the number of fencers who do not use the foil as their main weapon
  • 0, 1, 2, 3,... 25
  • X ~ B (25,0.40)
  • The probability that all 25 not use the foil is almost zero. Therefore, it would be very surprising.
  • X = the number of audits in a 20-year period
  • 0, 1, 2, …, 20
  • X ~ B (20, 0.02)
  • X = the number of matches
  • X ~ B ( 3 , 1 6 ) ( 3 , 1 6 )
  • In dollars: −1, 1, 2, 3
  • Multiply each Y value by the corresponding X probability from the PDF table. The answer is −0.0787. You lose about eight cents, on average, per game.
  • The house has the advantage.
  • Mean = μ = np = 15(0.281) = 4.215
  • Standard Deviation = σ = n p q n p q = 15 ( 0.281 ) ( 0.719 ) 15 ( 0.281 ) ( 0.719 ) = 1.7409
  • P ( x > 5) = 1 – P ( x ≤ 5) = 1 – binomcdf(15, 0.281, 5) = 1 – 0.7754 = 0.2246 P ( x = 3) = binompdf(15, 0.281, 3) = 0.1927 P ( x = 4) = binompdf(15, 0.281, 4) = 0.2259 It is more likely that four people are literate that three people are.
  • X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
  • X ~ G (0.40)
  • X = the number of pages that advertise footwear
  • X takes on the values 0, 1, 2, ..., 20
  • X ~ B (20, 29 192 29 192 )
  • X = the number of pages we must survey until we find one that advertises footwear. X ~ G ( 29 192 29 192 )
  • 6.6207 pages

0, 1, 2, and 3

  • X ~ G (0.25)
  • Mean = μ = 1 p 1 p = 1 0.25 1 0.25 = 4
  • Standard Deviation = σ = 1 − p p 2 1 − p p 2 = 1 − 0 .25 0.25 2 1 − 0 .25 0.25 2 ≈ 3.4641
  • P ( x = 10) = geometpdf(0.25, 10) = 0.0188
  • P ( x = 20) = geometpdf(0.25, 20) = 0.0011
  • P ( x ≤ 5) = geometcdf(0.25, 5) = 0.7627
  • 0, 1, 2, 3, ..., 20
  • X ~ H (29, 163, 20); r = 29, b = 163, n = 20
  • X = the number of Patriots picked
  • 0, 1, 2, 3, 4
  • X ~ H (4, 8, 9)
  • Without replacement
  • X ~ P (5.5); μ = 5.5; σ  =  5.5 σ  =  5.5 ≈ 2.3452
  • P ( x ≤ 6) = poissoncdf(5.5, 6) ≈ 0.6860
  • There is a 15.7% probability that the law staff will receive more calls than they can handle.
  • P ( x > 8) = 1 – P ( x ≤ 8) = 1 – poissoncdf(5.5, 8) ≈ 1 – 0.8944 = 0.1056

Let X = the number of defective bulbs in a string.

Using the Poisson distribution:

  • μ = np = 100(0.03) = 3
  • P ( x ≤ 4) = poissoncdf(3, 4) ≈ 0.8153

Using the binomial distribution:

  • X ~ B (100, 0.03)
  • P ( x ≤ 4) = binomcdf(100, 0.03, 4) ≈ 0.8179

The Poisson approximation is very good—the difference between the probabilities is only 0.0026.

  • X = the number of children for a Spanish woman
  • 0, 1, 2, 3,...
  • X ~ P (1.47)
  • X = the number of fortune cookies that have an extra fortune
  • 0, 1, 2, 3,... 144
  • X ~ B (144, 0.03) or P (4.32)
  • 0.0124 or 0.0133
  • 0.6300 or 0.6264
  • As n gets larger, the probabilities get closer together.
  • X = the number of people audited in one year
  • 0, 1, 2, ..., 100
  • X = the number of shell pieces in one cake
  • X ~ P (1.5)

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  • Authors: Barbara Illowsky, Susan Dean
  • Publisher/website: OpenStax
  • Book title: Introductory Statistics
  • Publication date: Sep 19, 2013
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/introductory-statistics/pages/1-introduction
  • Section URL: https://openstax.org/books/introductory-statistics/pages/4-solutions

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